0

我正在使用 username 和 device token 将设备注册到 stackmob 。我从 c2dn 获取有效令牌，然后将其存储到该用户的数据库中，然后在注册到 stackmob 时我正在使用这些参数。在开发环境中，它工作正常，但同一段代码在注册设备时给出 401。请在这方面给我建议。

代码如下：

public String registerWithNotificationServiceProvider(final String userName, final String deviceToken)        throws UserException 

         {
    if (userName.isEmpty() || deviceToken.isEmpty()) {
        throw new UserException(ResponseCodes.STATUS_BAD_REQUEST, "User Name or device      token is null",## Heading ##                    "label.invalid.user.device.details");
    }
    StackMobRequestSendResult deviceRegisterResult = null;
    deviceRegisterResult = StackMob.getStackMob().registerForPushWithUser(userName, deviceToken,
            new StackMobRawCallback() {
                @Override
                public void done(HttpVerb requestVerb, String requestURL,
                        List<Map.Entry<String, String>> requestHeaders, String requestBody,
                        Integer responseStatusCode, List<Map.Entry<String, String>> responseHeaders,
                        byte[] responseBody) {
                    String response = new String(responseBody);
                    logger.info("request Body is " + requestBody);
                    logger.info("request Url is " + requestURL);
                    for(Map.Entry<String, String> entry : requestHeaders){
                        logger.info("Request Header is " + entry.getKey());
                        logger.info("Request Header content is " + entry.getValue());
                    }
                    for(Map.Entry<String, String> entry : responseHeaders){
                        logger.info("Response Header is " + entry.getKey());
                        logger.info("Response Header content is " + entry.getValue());
                    }
                    logger.info("response while  registering the device is  " + response);
                    logger.info("responseCode while registering device " + responseStatusCode);
                }
            });
    String status = null;
    if (deviceRegisterResult.getStatus() != null) {
        status = deviceRegisterResult.getStatus().name();
        logger.debug("For user : " + userName + " Status for registering device is " + status);
    }
    if (Status.SENT.getStatus().equalsIgnoreCase(status)) {
        return Status.SUCCESS.getStatus();
    } else {
        return Status.FAILURE.getStatus();
    }

}
4

1 回答 1

1

当您使用 api 密钥和秘密设置 StackMob 对象时，您是否记得将 apiVersion 1 与您的生产密钥和秘密一起使用？这是最有可能的问题。

StackMobCommon.API_KEY = KEY;
StackMobCommon.API_SECRET = SECRET;
StackMobCommon.USER_OBJECT_NAME = "user";
StackMobCommon.API_VERSION = 1; //! 0 for dev, 1 for production

如果这不起作用还设置 StackMob.getStackMob().getLogger().setLogging(true); 在开头并发布生成的日志

于 2012-07-06T19:04:12.633 回答